Problem 12 - A Course in Calculus and Real Analysis

Question

Evaluate the following limits. (i) \(\lim _{h \rightarrow 0} \frac{1}{h} \int_{x}^{x+h} \frac{d u}{u+\sqrt{u^{2}+1}}\), (ii) \(\lim _{x \rightarrow 0} \frac{1}{x^{3}} \int_{0}^{x} \frac{t^{2} d t}{t^{4}+1}\), (iii) \(\lim _{x \rightarrow 0} \frac{1}{x^{6}} \int_{0}^{x^{2}} \frac{t^{2} d t}{t^{6}+1}\), (iv) \(\lim _{x \rightarrow x_{0}} \frac{x}{x-x_{0}} \int_{x_{0}}^{x} f(t) d t\) (v) \(\lim _{x \rightarrow x_{0}} \frac{x}{x^{2}-x_{0}^{2}} \int_{x_{0}}^{x} f(t) d t\), provided \(f\) is continuous at \(x_{0}\).

Step-by-Step Solution

Answer

(i) The limit is \(\frac{1}{x+\sqrt{x^{2}+1}}\). (ii) The limit is \(0\). (iii) The limit is \(0\). (iv) The limit is \(f(x_0)\). (v) The limit is \(\frac{f(x_0)}{2x_0}\).

Step 1: Finding the antiderivative

We begin by finding an antiderivative of \(\frac{1}{u+\sqrt{u^{2}+1}}\). The derivative of \(\log (u + \sqrt{u^2 + 1})\) is \(\frac{1}{u + \sqrt{u^2 + 1}}\), so we can write the integral as: $$ \frac{1}{h}\int_{x}^{x + h} \frac{1}{u + \sqrt{u^2 + 1}} du = \frac{1}{h}[\log (u + \sqrt{u^2 + 1})]_{x}^{x + h} $$

Step 2: Evaluate the limit

Now plug in the limits of integration and compute the limit as \(h \rightarrow 0\): $$ \lim_{h\rightarrow 0} \frac{1}{h}\left[\log\left((x+h)+\sqrt{(x+h)^{2}+1}\right)-\log\left(x+\sqrt{x^{2}+1}\right)\right] $$ Use the properties of logarithms to combine the difference of logs into a single log: $$ \lim_{h\rightarrow 0} \frac{1}{h}\log\frac{(x+h)+\sqrt{(x+h)^{2}+1}}{x+\sqrt{x^{2}+1}} $$ Now apply L'Hopital's rule: $$ \lim_{h\rightarrow 0} \frac{\frac{d}{dh}\log\left(\frac{(x+h)+\sqrt{(x+h)^{2}+1}}{x+\sqrt{x^{2}+1}}\right)}{1} = \frac{d}{dx}\log(x+\sqrt{x^{2}+1}) $$ The limit is equal to the derivative of \(\log(x+\sqrt{x^{2}+1})\) which is \(\frac{1}{x+\sqrt{x^{2}+1}}\). (ii)

Step 1: Finding the antiderivative

We start by finding an antiderivative of \(\frac{t^2}{t^4 + 1}\). The function doesn't have a simple antiderivative, but since we will be applying L'Hopital's rule, we can treat it as a single integral: $$ \int_{0}^{x} \frac{t^2}{t^4 + 1} dt $$

Step 2: Evaluate the limit

Now we'll directly apply L'Hopital's rule to the limit: $$ \lim_{x\rightarrow 0} \frac{1}{x^3}\int_{0}^{x} \frac{t^2}{t^4 + 1} dt = \lim_{x\rightarrow 0} \frac{\frac{d}{dx}\left(\int_{0}^{x} \frac{t^2}{t^4 + 1} dt\right)}{3x^2} $$ By the Fundamental Theorem of Calculus, the derivative of the integral is the original function, so our limit is: $$ \lim_{x\rightarrow 0} \frac{\frac{x^2}{x^4 + 1}}{3x^2} $$ After canceling the \(x^2\) terms and plugging in \(x = 0\), we have: $$ \frac{0}{3} = 0 $$ (iii)

Step 1: Finding the antiderivative

We similarly start by treating the integral as a single expression involving \(x^2\): $$ \int_{0}^{x^2} \frac{t^2}{t^6 + 1} dt $$

Step 2: Evaluate the limit

Now apply L'Hopital's rule: $$ \lim_{x\rightarrow 0} \frac{1}{x^6}\int_{0}^{x^2} \frac{t^2}{t^6 + 1} dt = \lim_{x\rightarrow 0} \frac{\frac{d}{dx}\left(\int_{0}^{x^2} \frac{t^2}{t^6 + 1} dt\right)}{6x^5} $$ The derivative of the integral is \(\frac{(x^2)^2}{(x^2)^6 + 1} = \frac{x^4}{x^{12} + 1}\), so our limit is: $$ \lim_{x\rightarrow 0} \frac{\frac{x^4}{x^{12} + 1}}{6x^5} $$ Cancel one of the \(x^4\) terms and notice that plugging in \(x = 0\) does not give an indeterminate form: $$ \frac{0}{6} = 0 $$ (iv)

Step 1: Recognize the form of the limit

This limit is also in the form \(\lim_{x \rightarrow c} \frac{F(x)}{x - c}\), where \(F(x) = \int_{x_0}^{x} f(t) dt\). By the Fundamental Theorem of Calculus, we know that \(F'(x) = f(x)\). This suggests applying L'Hopital's rule to the limit.

Step 2: Applying L'Hopital's rule

Applying L'Hopital's rule, we obtain: $$ \lim_{x\rightarrow x_0} \frac{x}{x - x_0}\int_{x_0}^{x}f(t) dt = \lim_{x\rightarrow x_0}\frac{f(x)}{1} = f(x_0) $$ (v)

Step 1: Recognize the form of the limit

This limit is also of the form \(\lim_{x \rightarrow c} \frac{F(x)}{x - c}\), with \(F(x) = \int_{x_0}^{x} f(t) dt\).

Step 2: Applying L'Hopital's rule

First, we rewrite the limit: $$ \lim_{x\rightarrow x_0} \frac{x}{x^2 - x_0^2}\int_{x_0}^{x}f(t) dt = \lim_{x\rightarrow x_0}\frac{x}{(x - x_0)(x + x_0)}\int_{x_0}^{x}f(t) dt $$ Applying L'Hopital's rule, we obtain: $$ \lim_{x\rightarrow x_0}\frac{f(x)}{2x} = \frac{f(x_0)}{2x_0} $$

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a central idea in calculus and connects the concept of differentiation with that of integration. It essentially states that if you take an antiderivative of a function and then differentiate it, you return to the original function. This is expressed as follows: if \ f \ is continuous over the interval \ [a, b] \ and \ F \ is an antiderivative of \ f \, then:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)
• \( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \)

Through this theorem, we can evaluate definite integrals and find the accumulated change easily. In the exercises mentioned, this theorem helps in transforming the problem of evaluating an integral into differentiating an antiderivative, making calculations much easier.

L'Hopital's Rule

L'Hopital's Rule is a very useful tool in calculus for evaluating limits of indeterminate forms. When a limit falls into the form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hopital's Rule guides you to differentiate the numerator and the denominator separately, and then find the limit again. This process can be repeated as necessary, assuming the limit still remains in an indeterminate form after taking derivatives.
To apply L'Hopital's Rule correctly:

• Ensure your limit is of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiate the top (numerator) and the bottom (denominator) separately.
• Recalculate the limit of the fraction of derivatives.

In the context of the given exercise problems, the rule helps in simplifying complex integral expressions when plugged into the limit, turning them into a form that's easier to handle. L'Hopital's Rule allows bypassing tedious algebraic manipulations by focusing on derivatives, driving directly to the solution.

Antiderivatives

Before discussing antiderivatives, it's important to understand that they are the reverse operation of derivatives. If the derivative of a function \( f(t) \) is known, an antiderivative \( F(t) \) is a function such that \ \frac{d}{dt} F(t) = f(t) \.
Finding an antiderivative means you're basically solving the problem of integration, albeit one without specific bounds (unlike definite integrals).

• Antiderivatives provide a way to evaluate the potential function from its derivative.
• When solving problems involving limits and integrals, recognizing antiderivatives can simplify the process significantly, as seen in the exercises covered.

Antiderivatives appear in the solutions as part of the method to simplify the integrals before applying the Fundamental Theorem of Calculus or L'Hopital's Rule.

Continuity

The concept of continuity is crucial in calculus because it ensures that function values change smoothly without any jumps, breaks, or holes. Specifically, a function \( f \) is continuous at a point \( x_0 \) if the following three conditions hold:

• \( f(x_0) \) is defined,
• \( \lim_{x \to x_0} f(x) \) exists,
• \( \lim_{x \to x_0} f(x) = f(x_0) \).

In contexts involving calculus, especially when employing the Fundamental Theorem of Calculus, continuity over the interval is essential for ensuring that integrals behave predictably.
In the provided problems, the assumption of continuity allows for applying the Fundamental Theorem of Calculus without any adjustments for discontinuous points, ensuring the integration and differentiation processes are seamless and consistent. Hence, continuity plays a key role in simplifying and solving calculus problems effectively.